leetcode64. minimum path and

Given a non-negative integer of mxn grid, find a path from left to bottom right, so that the sum of the minimum number of paths.

Note: you can only move one step down or to the right.

Example:

输入:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
输出: 7
解释: 因为路径 1→3→1→1→1 的总和最小。

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/minimum-path-sum
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The complete code

Using the results of running overtime thinking back, a total of 61 test cases, only 60 passed.
Despite all conceivable pruning shears, so I really can not think of how to optimize.

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        if(grid.size() == 0)
            return 0;
        int res = INT_MAX;
        path(grid, res, 0, 0, 0);
        return res;        
    }
private:
    void path(vector<vector<int>>& grid, int &res, int cur, int i, int j){        
        if(i == grid.size() - 1 && j == grid[0].size() - 1){
            cur += grid[i][j];
            if(cur < res)
                res = cur;
            return;
        }
        int temp;
        temp = cur + grid[i][j];
        //向下走
        if(i + 1 < grid.size() && temp < res)
            path(grid, res, temp, i + 1, j);
        //向右走
        if(j + 1 < grid[0].size() && temp < res)
            path(grid, res, temp, i, j+1);
    }
};

Dynamic Programming
In fact, this question is a typical topic of dynamic programming, backtracking blame too familiar with the

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        if(grid.size() == 0)
            return 0;
        int res = INT_MAX;
        path(grid, res, 0, 0, 0);
        return res;        
    }
private:
    void path(vector<vector<int>>& grid, int &res, int cur, int i, int j){        
        if(i == grid.size() - 1 && j == grid[0].size() - 1){
            cur += grid[i][j];
            if(cur < res)
                res = cur;
            return;
        }
        int temp;
        temp = cur + grid[i][j];
        //向下走
        if(i + 1 < grid.size() && temp < res)
            path(grid, res, temp, i + 1, j);
        //向右走
        if(j + 1 < grid[0].size() && temp < res)
            path(grid, res, temp, i, j+1);
    }
};

执行结果：通过
显示详情
执行用时 :12 ms, 在所有 C++ 提交中击败了57.07%的用户
内存消耗 :11 MB, 在所有 C++ 提交中击败了45.90%的用户